%I #15 Feb 18 2015 18:11:06

%S 1,2,7,34,208,1504,12283,109778,1050820,10614856,111978128,1224261856,

%T 13792583296,159411938560,1883550536707,22687603653106,

%U 277940485660012,3456490397570392,43565433620294908,555752354850506312,7167182317486700416,93348781597357983232,1226830676118851157712

%N Number of permutations of length 2n+1 which are invariant under the reverse-complement map and have no decreasing subsequences of length 5.

%F a(n) = sum(j=0, n, C(n,j)^2 * A005802(j)).

%F a(n) = sum(j=0, n, C(n,j)^2 * (1/((j+1)^2 (j+2))) * sum(i=0, j, C(2*i,i) * C(j+1,i+i) * C(j+2,i+1))) where C(n,j) = n!/(j!(n-j)!).

%F Recurrence: (n+2)^3*(3*n+1)*a(n) = 2*(30*n^4 + 67*n^3 + 29*n^2 - 10*n - 8)*a(n-1) - 64*(n-1)^2*n*(3*n+4)*a(n-2). - _Vaclav Kotesovec_, Feb 18 2015

%F a(n) ~ 2^(4*n+5) / (Pi^(3/2) * n^(9/2)). - _Vaclav Kotesovec_, Feb 18 2015

%t Table[Sum[ Binomial[n, j]^2*(1/((j + 1)^2*(j + 2)))* Sum[Binomial[2*i, i]*Binomial[j + 1, i + 1]* Binomial[j + 2, i + 1], {i, 0, j}], {j, 0, n}], {n, 0, 20}]

%o (PARI) /* using formula given; this gives fractions! */

%o C=binomial;

%o a(n)=sum(j=0, n, C(n,j)^2 * (1/((j+1)^2*(j+2))) * sum(i=0, j, C(2*i,i)*C(j+1,i+i)*C(j+2,i+1)));

%o \\ _Joerg Arndt_, Feb 18 2015

%o (PARI) /* Using a(n) = sum(j=0, n, C(n,j)^2 * A005802(j)). */

%o f(n)= 2 * sum(k=0,n, binomial(2*k, k) * (binomial(n, k))^2 * (3*k^2+2*k+1-n-2*k*n)/((k+1)^2 * (k+2) * (n-k+1)));

%o vector(33, N, my(n=N-1); sum(j=0,n, f(j) * C(n,j)^2 ) )

%o \\ _Joerg Arndt_, Feb 18 2015

%K nonn

%O 0,2

%A _Eric S. Egge_, Oct 21 2008

%E Added more terms, _Joerg Arndt_, Feb 18 2015